Chapter 4
Types of Reactions &
Solution Stoichiometry
1
UNIT ESSENTIAL QUESTION:
How do chemicals react with one another
in aqueous solutions?
2
LESSON ESSENTIAL
QUESTIONS (4.1-4.4):
1) How do water molecules interact with
chemicals?
2) How is the concentration of a solution
measured?
3
Section 4.1
Water, the Common Solvent
4
Aqueous solutions
Dissolved in water.
Good solvent- polar molecules.
Hydration: ions in salts break
apart due to attraction to polar
water molecules.
Example:
NH4NO3 (s)  NH4+ (aq) + NO3- (aq)
δ+
δ-
δ+
5
Hydration
H
H
6
H
H
H
Solubility
Amount of substance that will dissolve in
a given amount of water.
If they do dissolve, ions are separated,
and can move around.
Water can also dissolve non-ionic
compounds if they have polar bonds.
7
“Like dissolves like”
Polar substances generally dissolve
other polar and ionic substances
– Alcohol is slightly polar and dissolves
(mixes) in water
Nonpolar substances dissolve other
nonpolar substances
– Fat will not dissolve in water
8
Section 4.2
The Nature of Aqueous Solutions:
Strong & Weak Electrolytes
9
Parts of Solutions
Solution- homogeneous mixture.
Solute- what gets dissolved.
Solvent- what does the dissolving.
Soluble- Able to be dissolved.
Miscible- liquids dissolve in each other.
10
Electrolytes
Electrolytes- ionic compounds in solution that
conduct electricity.
Strong electrolytes- completely dissociate (fall
apart into ions).
– Many ions = conduct electricity well.
Weak electrolytes- partially dissociate into
ions.
– Few ions = conduct electricity slightly.
Non-electrolytes- don’t dissociate at all.
– No ions = don’t conduct electricity.
11
Acid/Base Electrolytes
Arrhenius acid- forms H+ ions when dissolved.
Strong acids dissociate completely.
– Ex: H2SO4 HNO3 HCl HBr HI
Weak acids do not dissociate completely.
– Ex: HC2H3O2
Arrhenius base - forms OH- ions when
dissolved.
Strong bases also dissociate completely.
– Ex: KOH NaOH (Groups 1 & 2
hydroxides)
12
Sections 1&2 Homework
Pg. 170-171 #1,9,18,19
13
Warm-Up
HNO3 is a strong acid. Write
the chemical equation for a
solution of HNO3. Will it
conduct electricity?
14
Section 4.3
Composition of Solutions
15
Measuring Composition of Solutions
To do stoichiometry:
– Need to know chemicals
– Need to know amounts (concentrations)
Concentration- how much is dissolved.
Molarity = Moles of solute
Liters of solution
abbreviated M (molar)
1 M = 1mol solute / 1 liter solution
16
Molarity Calculations
Can solve for:
– Amount or mass of solid to dissolve
– Moles of solute
– Volume of solution
– Standard solution
• Solution whose concentration is accurately
known.
17
Examples
Calculate the molarity of a solution
prepared by dissolving 11.5g of solid
NaOH in water to make 1.50L of
solution. (pg. 134)
Give the concentration of each ion in
0.50 M Co(NO3)2. (pg. 135)
#27 pg. 172
18
Dilutions
Stock solution – a concentrated solution
Dilution – number of moles of solute
stays the same, just adding more water
– M1V1 = M2V2 mol1 x V1 = mol2 x V2
L1
Example: #30 (a) pg. 172
19
L2
Section 3 Homework
Pg. 171-172 #21-23,28,31
20
LESSON ESSENTIAL
QUESTIONS (4.5-4.7):
1) How do we identify and work with
precipitation reactions?
21
Section 4.5
Precipitation Reactions
22
23
Precipitation Reactions
Solid forms when two solutions of ionic
compounds are mixed.
Precipitate (ppt)
To help you remember: ‘If you’re not a
part of the solution, your part of the
precipitate!’
24
Precipitation reactions
NaOH(aq) + FeCl3(aq) 
NaCl(aq) + Fe(OH)3(s)
is really:
Na+(aq)+OH-(aq) + Fe+3 + Cl-(aq) 
Na+ (aq) + Cl- (aq) + Fe(OH)3(s)
So all that really happens is
OH- (aq) + Fe+3 (aq)  Fe(OH)3 (s)
net ionic equation!
Also a double displacement reaction
25
Precipitation reaction
Can predict products, but can only be certain
by experimenting.
The anion and cation switch partners.
Only occurs if a product is insoluble!
– Otherwise all the ions stay in solutionnothing has happened (spectators)
Memorize solubility rules! Pg. 144
26
Solubility Rules
 All nitrates, Na+, K+, NH4+ are soluble.
You must know this for the AP exam!
Additional solubility rules on pg. 144.
27
Section 4.6
Describing Reactions in Solutions
28
Three Types of Equations
1. Formula Equation- write formulas, not
ions.
K2CrO4(aq) + Ba(NO3)2(aq) 
2. Complete Ionic equation- show
dissolved electrolytes as the ions.
2K+ + CrO4-2 + Ba+2 + 2 NO3- 
BaCrO4(s) + 2K+ + 2 NO3Spectator ions are those that don’t
react- appear as ions on both sides.
29
Three Type of Equations
3. Net Ionic equation- show only ions
that react, not spectator ions
Ba+2 + CrO4-2  BaCrO4(s)
If all species in a reaction are aqueous
(soluble), write NR!
30
Sections 5&6 Homework
Pg. 172-173 #36,42,44
31
AP Practice Question
How many moles of Na2SO4 must be added to 500
milliliters of water to create a solution that has a 2molar concentration of the Na+ ion? (Assume the
volume of the solution does not change.)
 Think about what you need to answer this!
 Need to find moles Na+. Then find moles Na2SO4
a)
b)
c)
d)
32
0.5 moles
1 mole
2 moles
5 moles
Section 4.7
Stoichiometry of Precipitation
Reactions
33
Stoichiometry of Precipitation
Steps for reference: pg.148
– Similar to other stoichiometry problems
we’ve done!
Sample problem: What volume of 0.15M
KCl is needed to precipitate out all of the
lead ions from 100.mL of 0.20M
Pb(NO3)2?
270mL KCl needed
34
Section 7 Homework
Pg. 173 #47,48,50,54
35
Section 4.8
Acid-Base Reactions
36
LESSON ESSENTIAL
QUESTION (4.8):
1)
2)
37
How do we classify acids and bases?
What happens when acids and bases
are mixed together?
Acid-Base Reactions
For our purposes an acid is a proton donor, H+
(BrØnsted-Lowry theory).
A base is a proton acceptor, usually OHacid + base  salt + water
H+ + OH-  H2O
Practice: Write the net ionic equation for the
acid/base rxn. below:
HNO3(aq) + NaOH(aq)  ?
Note: H2CO3 always breaks down into CO2 &
H2O
38
Acid-Base Reactions
Follow same steps as precipitation
reactions for stoichiometry problems.
– See p.149-150
Practice: What volume (in mL) of 0.100M
HCl will react completely with 25.00mL
of 0.200 M NaOH?
– (1) Write net ionic equation
– (2) Find moles you’re starting with
– (3) Find moles needed
– (4) Find volume needed
39
Acid-Base Reactions
Also called neutralization reactions.
Use titrations to determine concentrations.
Titrant: solution of known concentration
Analyte: solution of unknown concentration
Equivalence Point: when enough titrant has been
added to exactly react with the analyte
(neutralization is complete).
– Stoichiometric amounts come from balanced equation!
– Tells us how many moles of the titrant fully reacted
with the analyte- then can solve for moles of analyte!
40
Titration
Solution of known concentration (titrant),
is added to the unknown (analyte), until
the equivalence point is reached.
How do we know when the equivalence
point has been reached?
– Add indicator to analyte at the beginning
41
Titration
Where the indicator changes color is the
endpoint.
– Ex: phenolphthalein used often
• Pink in base, colorless in acid
As close as we can get to the equivalence
point; still assume complete neutralization.
– The solution will not turn pink until one
drop after the equivalence point (when the
solution is more basic).
Can also use titration for non acid/base
substances to find amounts/concentrations.
42
AP Practice Question
Which of the following best represents the
balanced net ionic equation for the reaction of
lead(II) carbonate & concentrated hydrochloric
acid? (All lead compounds are insoluble.)
a.
b.
c.
d.
43
Pb2CO3 + 2H+ + Cl-  Pb2Cl + CO2 + H2O
PbCO3 + 2H+ + 2Cl-  PbCl2 + CO2 + H2O
PbCO3 + 2H+  Pb+2 + CO2 + H2O
PbCO3 + 2Cl-  PbCl2 + CO3-2
AP Practice Question
The conductivity of a solution of Ba(OH)2 is
monitored as the solution is titrated with 0.10 M
H2SO4. The original volume of the Ba(OH)2 solution
is 25.0 mL. A precipitate of BaSO4 is formed during
the titration. The data collected from the experiment
is plotted in the graph above.
44
Question Continued
1) As the first 30.0 mL of 0.10 M H2SO4 are added to
the Ba(OH)2 solution, two types of chemical
reactions occur simultaneously. Write the
balanced net-ionic equations for (i) the
neutralization reaction and (ii) the precipitation
reaction.
(i) Equation
for+neutralization
OH- (aq)
2H+ (aq)  H reaction:
O (l)
2
(ii) Equation
for+precipitation
Ba+2 (aq)
SO -2 (aq)  reaction:
BaSO (s)
4
45
4
Question Continued
2) The conductivity of the Ba(OH)2 solution decreases as the
volume of added 0.10 M H2SO4 changes from 0.0 mL to 30.0
mL.
(i) Identify the chemical species that enable the solution to
conduct electricity as the first 30.0 mL of 0.10 M H2SO4 are
added. OH- (aq) & Ba+2 (aq) (Can’t be anything from
H2SO4 because the ions immediately react.)
(ii) On the basis of the equations you wrote in question 1,
explain why the conductivity decreases.
[Ba+2] in sltn. decrease as they precipitate out, and [OH-] in
sltn. decrease as they react to form H2O. Note: be specific in
your answers!! Reference all species and reactions!
46
Question Continued
Think about what
information can be
determined from this
point!
At equivalence point:
complete neutralization
3) Using the information in the graph, calculate the molarity
of the original Ba(OH)2 solution. 0.12M Ba(OH)2
47
Section 8 Homework
Homework: pg.173-174
#56, 58, 60, 64, 66
48
Section 4.9
Oxidation – Reduction Reactions
49
LESSON ESSENTIAL
QUESTIONS (4.9-4.10):
1)
2)
3)
50
How can we identify redox reactions?
How do we assign oxidation states?
Why is balancing different for redox
reactions?
51
Redox Reactions
Ionic compounds are formed through
the transfer of electrons.
An oxidation-reduction reaction involves
the transfer of electrons.
– One element gains, one loses
Non-ionic compounds can also undergo
redox reactions.
52
Oxidation States = ‘charge’
A way of keeping track of the electrons.
Not necessarily true of what is in nature, but
it works.
Need to memorize rules for assigning
(pg.156):
 The oxidation state of elements in their
standard states is zero.
 Oxidation state for monatomic ions are the
same as their charge.
53
Oxidation states
 Oxygen is assigned an oxidation state of -2
in its covalent compounds except in
peroxide (-1).
 In compounds with nonmetals hydrogen is
assigned the oxidation state +1.
 In its compounds fluorine is always –1.
 The sum of the oxidation states must be
zero in compounds or equal the charge of
the ion.
54
Practicing Oxidation States
Determine the oxidation states in the
following:
1)
2)
3)
4)
55
Cl2 Cl: 0
O: -2
SO4-2 S: +6
CaBr2 Ca: +2 Br: -1
C6H12O6 C: 0 H: +1 O: -2
Section 9 Homework
Pg. 174 #67(c-e),68(a-c),72
56
Section 4.10
Balancing Redox Reactions
57
Oxidation-Reduction
e- transferred, so the oxidation states
change.
Oxidation is the loss of electrons.
– More positive oxidation state.
Reduction is the gain of electrons.
– More negative oxidation state.
OIL RIG
LEO (the lion goes) GER
58
Agents
Oxidizing agent- substance that gets
reduced (causes oxidation in another
species).
– Gains electrons.
– More negative oxidation state.
Reducing agent- substance that gets
oxidized (causes reduction in another
species).
– Loses electrons.
– More positive oxidation state.
59
Identify the…
Oxidizing agent
Reducing agent
Substance oxidized
Substance reduced
reducing agent,
substance oxidized
oxidizing agent,
substance reduced
#1: 2Na + Cl2  2NaCl
reducing agent,
substance oxidized
oxidizing agent,
substance reduced
#2: CH4 + 2O2  CO2 + 2H2O
60
Half-Reactions
All redox reactions can be thought of as
happening in two halves.
One produces electrons - oxidation half.
The other requires electrons - reduction
half.
Ex: Fe (s) + CuSO4 (aq)  Cu (s) + FeSO4 (aq)
Net Ionic: Fe (s) + Cu+2 (aq)  Cu (s) + Fe+2 (aq)
Oxidation: Fe (s)  Fe+2 (aq) + 2eReduction: Cu+2 (aq) + 2e-  Cu (s)
61
Balancing Redox Equations
Redox reactions may involve an acid or
base as a reactant.
The number of electrons produced
must be the same as those required.
8 step procedure for acidic solution, 10
step procedure for basic solution.
Called the half reaction method.
– Balance each half reaction, then
combine for total balanced reaction
62
Balancing in Acidic Solution
 Write separate half reactions.
 For each half reaction balance all
species except H and O.
 Balance O by adding H2O to one side.
 Balance H by adding H+ to one side.
 Balance charge by adding e- to the
more positive side.
63
Balancing in Acidic Solution
 Multiply equations by a number to make
electrons equal.
 Add equations together and cancel
identical species. Reduce coefficients to
smallest whole numbers.
 Check that charges and elements are
balanced.
64
Balancing in Acidic Solution
Ex: Balance the following equation:
H+ (aq) + Cr2O7-2 (aq) + C2H5OH (l) 
Cr+3 (aq) + CO2 (g) + H2O (l)
Reduction: 6e- + 14H+ + Cr2O7-2  2Cr+3 + 7H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+ + 12eFinal: 16H+ + 2Cr2O7-2 + C2H5OH  4Cr+3 +
11H2O + 2CO2
*Note: there should NOT be any e- in the final balanced
equation! If so, not balanced!
65
Basic Solution
Do everything you would with acid, but
add one more step.
Add enough OH- to both sides to
neutralize the H+.
Any H+ and OH- on the same side form
water. Cancel out any H2O’s on both
sides.
Simplify coefficients, if necessary.
66
Balancing in Basic Solution
Assume previous example in acidic solution was
actually in a basic solution.
Had: 16H+ + 2Cr2O7-2 + C2H5OH  4Cr+3 + 11H2O
+ 2CO2
For any H+ ions, add same number of OH- ions
to both sides. This forms water with H+. Cancel
out waters on both sides. 16H+, so add 16OH5 H2 O
Now: 16H2O + 2Cr2O7-2 + C2H5OH  4Cr+3 +
11H2O + 2CO2 + 16OH67
Practice Balancing Redox Rxns.
Pg. 174 #74(b)
Answer: 6Cl- + Cr2O7 + 14H+  3Cl2 + 2Cr+3 + 7H2O
Pg. 174 #75(b)
Answer: 2OH- + Cl2  OCl- + Cl- + H2O
68
Side Note: Redox Titrations
Same as titrations discussed before, just
looking at redox reactions instead of
acid/base reactions.
Permanganate ion is used often because it
is its own indicator: MnO4- is purple, Mn+2
is colorless. When reaction solution
remains clear, MnO4- is gone.
Chromate ion is also useful, but color
change, orangish yellow to green, is
harder to detect.
69
Section 10 Homework
Pg. 174-175 #73-76 ONLY letter
a for each
70